Question 13 Marks
4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.How many moles of HCl are used in this reaction
AnswerMolecular mass of HCl=36.5
Molecular mass of calcium carbonate =100
As 100g of calcium carbonate gives (2 × 36.5)= 73g of HCl,
∴ 450g of calcium carbonate will give $\frac{450 \times 73}{100}$=328.5g
Number of moles of HCl = $\begin{aligned} & \frac{\text { weight of } HCl }{\text { Molecular weight of } HCl } \\ & =\frac{328.5}{36.5}=9 \text { moles. }\end{aligned}$
View full question & answer→Question 23 Marks
4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid. What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111)
AnswerMolecular mass of calcium carbonate =100
Relative molecular mass of calcium chloride =111
As 100g of calcium carbonate gives 111g of calcium chloride.
∴ 450g of calcium carbonate will give $\frac{450 \times 111}{100}$ = 499.5g
View full question & answer→Question 33 Marks
$4.5 \ moles$ of calcium carbonate are reacted with dilute hydrochloric acid. What is the volume of carbon dioxide liberated at $\text{STP}$?
Answer$CaCO _3+2 \text{HCl} \longrightarrow CaCl _2+ H _2 O + CO _2 \uparrow$
As $100g$ of calcium carbonate gives $22.4 \ dm^3$ of $CO_{2,}$
$\therefore 450g$ of calcium carbonate will give
$\frac{450 \times 22.4}{100} = 100.8L$
View full question & answer→Question 43 Marks
4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.What is the mass of 4.5 moles of calcium carbonate? Relative molecular mass of calcium carbonate is 100)
AnswerRelative molecular mass of calcium carbonate=100
Mass of 4.5 moles of calcium carbonate
= No. of moles × Relative molecular mass
= 4.5×100
= 450g
View full question & answer→Question 53 Marks
Calculate the percentage of nitrogen and oxygen in ammonium nitrate.[Relative molecular mass of ammonium nitrate is 80, H=1, N=14, O = 16].
AnswerMolecular mass of $NH _4\left( NO _3\right)=80$
$
H =1, N =14, O =16
$
$\%$ of nitrogen
As $80 g$ of $NH _4\left( NO _3\right)$ contains $28 g$ of nitrogen,
$
\begin{aligned}
& \therefore 100 g \text { of } NH _4\left( NO _1\right) \text { will contain } \frac{28 \times 100}{80} \\
& =35 \%
\end{aligned}
$
$\%$ of nitrogen
As $80 g$ of $NH _4\left( NO _3\right)$ contains $48 g$ of oxygen
$
\begin{aligned}
& \therefore 100 g \text { of } NH _4\left( NO _1\right) \text { will contain } \frac{100 \times 48}{80} \\
& =60 \%
\end{aligned}
$
View full question & answer→Question 63 Marks
$\text{LPG}$ stands for liquefied petroleum gas. Varieties of $\text{LPG}$ are marketed including a mixture of propane $(60\%)$ and butane $(40\%)$. If $10$ litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reaction can be presented as :
$C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(g)}$
$2C_4H_{10(g) }+ 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(g)}$
Answer$10$ litres of $\text{LPG}$ contains
Propane $=\frac{60}{100} \times 10=6$ litres
Butane $=\frac{40}{100} \times 10=4$ litres
$ C _3 H _{8( g )}+5 O _{2( g )} \rightarrow 3 CO _{2( g )}+4 H _2 O _{( g )}$
$ 1 \text { Vol.(6L) } 3 \text { Vol.(18L) }$
$ 2 C _4 H _{10( g )}+13 O _{2( g )} \rightarrow 8 CO _{2( g )}+10 H _2 O _{( g )}$
$ 2 \ VOI \text { (4L) } 8\ Vol.$
$ (16 L )$
$ 18+16=34 L \text {. }$
$ $
View full question & answer→Question 73 Marks
Commercial sodium hydroxide weighing $30g$ has some sodium chloride in it. The mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing $14.3g$. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is:
$NaCl + AgNO_3\rightarrow AgCI + NaNO_3$
$[$Relative molecular mass $= 58, AgCI = 143]$
Answer$NaCl + AgNO_3\rightarrow AgCI + NaNO_3$
As $143g$ of $AgCl$ is obtained from $=58g$ of $NaCl$
So$, 14.3g$ of $AgCI$ will be obtained from $= 58 \times 14.3 / 143 = 5.8g$ of $NaCl$
Weight of commercial $NaOH = 30g$
Percentage of $NaCl$ in $NaOH = 5.8 \times 100 / 30 = 19.33\%$
View full question & answer→Question 83 Marks
Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with the moisture. During this reaction calcium hydroxide and acetylene gas is formed.
If $200\ cm^3$ of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion . The combustion reaction can be represented as below:
$2C_2H_{2(g)} + 5O_{2(g)} \rightarrow 4CO_{2(g)} + 2H_2O_{(g)}.$
AnswerApplying Gay $-$ Lussac's law on the equation :
$2C_2H_{2(g)} + 5O_{2(g)} \rightarrow 4CO_{2(g)} + 2H_2O_{(g)}.$
$2\ vol. 5\ vol. 4 \ vol.$
As $2$ volume of acetylene requires $= 5$ volume of oxygen
So$, 200 \ cm^3$ of acetylene will require $= 5 \times 200 / 2 = 500 \ cm^3$
Now further$, 2$ volume of acetylene produces $= 4$ volume of carbon dioxide
So$, 200 \ cm^3$ of acetylene will produce $= 4 \times 200 / 2 = 400 \ cm^3$
Hence$, 500 \ cm^3$ of oxygen and $400 \ cm^3$ of carbon dioxide is formed.
View full question & answer→Question 93 Marks
From the equation :
$C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2$
Calculate:
$(i)$ The mass of carbon oxidized by $49g$ of sulphuric acid $(C = 12,$ relative molecular mass of sulphuric acid $= 98)$
$(ii)$ The volume of sulphur dioxide measured at $\text{STP},$ liberated at the same time.
Answer$(i) \ C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2$
From the equation :
$2 \ moles$ of sulphuric acid oxidizes $1 \ mole$ of carbon
i.e. $2 \times 98g$ of sulphuric acid oxidizes $12g$ of carbon
So$, 49g$ of sulphuric acid will oxidize $= 12 \times 49 / 196 = 3g$
$3 g$ of carbon is oxidized by $49 g$ of sulphuric acid .
$(ii)$ Again from the equation :
$2\ moles$ of sulphuric acid liberates $2\ moles$ of sulphur dioxide.
i.e. $196g$ of sulphuric acid liberates $= 44.8\ dm^3$ of sulphur dioxide
$49g$ of sulphuric acid will liberate $= 44.8 \times 49 / 196 = 11.2 \ dm^3$
Hence$, 11.2\ dm^3$ of sulphur dioxide is liberated at the same time.
View full question & answer→Question 103 Marks
The equation for the burning of octane is :
$2C_8H_{18} + 25O_{2 }\rightarrow 16CO_2 + 18H_2O$
How many moles of carbon dioxide are produced when one mole of octane burns ?
Answer$2C_8H_{18} + 25O_2\rightarrow 16CO_2 + 18H_2O$
From the equation :
$2 \ moles$ of $C_8H_{18}$ produces $= 16\ moles$ of $CO_2$
$1 \ mole$ of $C_8H_{18 }$ will produce $= 16/2 = 8 \ moles$ of $CO_2$
So$, 8 \ moles$ of $CO_{2}$ is produced .
View full question & answer→Question 113 Marks
A compound $'X\ '$ consists of $4.8\%$ of $C$ and $95.2\%$ of Br by mass.
If the vapour density of the compound is $252,$ what is the molecular formula of the compound?
AnswerVapour density $= 252$
Empirical formula mass $= 12 + 3 \times 80 = 252$
Molecular mass $= 2 \ \times$ Vapour density
$= 2 \times 252 = 504$
Now, n $=$ Molecular mass / Empirical Formula Mass
$= 504/252 = 2$
Molecular formula $=$ n $\times$ Empirical Formula
$= 2 \times CBr_3 = C_2Br_6$
View full question & answer→Question 123 Marks
A compound $'X\ '$ consists of $4.8\%$ of $C$ and $95.2\%$ of Br by mass.
Determine the empirical formula of this compound.
$[C = 12, Br = 80]$
AnswerEmpirical formula of compound :
| Element |
Atomic mass |
Percentage |
Relative Number of moles |
Simplest mole ratio |
Whole number ratio |
| $C$ |
$12$ |
$4.8$ |
$4.8/12 = 0.4$ |
$0.4/0.4 = 1$ |
$1$ |
| $Br$ |
$80$ |
$95.2$ |
$95.2/80 = 1.19$ |
$1.19/0.4 = 3$ |
$3$ |
Empirical formula of the compound is $CBr_3.$ View full question & answer→Question 133 Marks
A sample of ammonium nitrate when heated yields $8.96$ litres of steam $($measured at $\text{STP})$
$NH_4NO_3 \rightarrow N_2O + 2H_2O$
What volume of dinitrogen oxide is produced at the same time as $8.96\ L$ of steam?
Answer$NH_4NO_3 \rightarrow N_2O + 2H_2O$
From the equation:
$1 \ mole$ of $NH_4NO_3$ yields $2\ mole$ of $H_2O.$
So$, 44.8L$ of steam $= 22.4L$ of $N_2O$ at $\text{STP}.$
$8.96L$ of Steam $= 22.4 \times 8.96/44.8 = 4.48L$ of $N_2O$ at $\text{STP}.$
View full question & answer→Question 143 Marks
$560ml$ of carbon monoxide is mixed with 500ml of oxygen and ignited. The chemicaI equation for the reaction is as folIows :
$2CO + O_2 \rightarrow 2CO_2$
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
Answer$2CO + O_2 \rightarrow 2CO_2$
From the reaction :
$2$ volumes of $CO$ consumes $= 1$ volume of $O_2$
So$, 560\ ml$ of $CO$ consumes $= 112 \times 560 = 280\ ml$
Now$, 2$ volume of $CO$ gives $= 2$ volume of $CO_2$
So$, 560\ ml$ of $CO$ will give $= 2 \times 560 / 2 = 560\ ml$
View full question & answer→Question 153 Marks
Calculate the number of moles and number of molecules present in $1.4 g$ of ethylene gas. What is the volume occupied by the same amount of ethylene?
AnswerMolecular mass of ethylene $(CH_2-CH_2) = 28g$
Number of $moles =$ Given weight/ Molecular weight
$= 1.4/28 = 0.05 \ moles$
Now, number of molecules in $1 \ mole = 6.023 \times 10^{23}$
So, number of molecules in $0.05 \ moles$
$= 6.023 \times 10^{23}\times 0.05 = 0.3 \times 10^{23}$
$= 3 \times 10^{22}$ molecules.
Volume occupied by $1\ mole$ of ethylene $= 22.4 L$
So, volume occupied by $0.05\ moles$ of ethylene
$= 22.4 \times 0.05 = 1.12 L$
View full question & answer→Question 163 Marks
Given that the relative molecular mass of copper oxide is $80,$ what volume of ammonia $($measured at $\text{STP})$ is required to completely reduce $120g$ of copper oxide? The equation for the reaction is :
$3CuO + 2NH_{3 }\rightarrow 3Cu + 3H_2O + N_2$
Answer$3CuO + 2NH_{3 }\rightarrow 3Cu + 3H_2O + N_2$
Molecular mass of $CuO = 80$
Volume occupied by $1 \ mole$ of $NH3 = 22.4L$
From the equation :
$3 \ moles$ of $CuO$ is reduced by $2\ moles$ of ammonia
For $240g$ of $CuO,$ volume of $NH_3$ consumed $= 44.8L$
For $120g$ of $CuO,$ volume of $NH_3$ consumed
=$ 44.8 \times 120 / 240 = 22.4 L$
View full question & answer→Question 173 Marks
Determine the empirical formula of a compound containing $47.9\% K, 5.5\%$ beryllium and $46.6\%$ fluorine by mass.
AnswerDetermination of empirical formula :
| Element |
Atomic mass |
Percentage |
Relative Number of moles |
Simplest mole ratio |
Whole number ratio |
| $K$ |
$39$ |
$47.9$ |
$47.9/39 =1.2$ |
$1.2/0.6 = 2$ |
$2$ |
| $Be$ |
$9$ |
$5.5$ |
$5.5/9 = 0.6$ |
$0.6/0.6 = 1$ |
$1 $ |
| $F$ |
$19$ |
$46.6$ |
$46.6/19 =2.4$ |
$2.4/0.6 =4$ |
$4$ |
The empirical formula of compound $= K_2BeF_{4.}$ View full question & answer→Question 183 Marks
Calculate the percentage of nitrogen in aluminium nitride. $[Al = 27, N = 14]$
AnswerMolecular mass of aluminium nitride $(\text{AIN}_3) = 27 + 14 \times 3 = 69$
Now$, 69$ parts by weight of aluminium nitride contains $= 42$ parts by weight of nitrogen
So$, 100$ parts will contain $= 42 \times 100 / 69 = 60.86\%$
Hence, the percentage of nitrogen in aluminium nitride is $60.86\%$
View full question & answer→Question 193 Marks
The volumes of gases $\text{A, B, C}$ and $\text{D}$ are in the ratio$, 1:2:2:4$ under the same conditions of temperature and pressure.
If the volume of $\text{A}$ is actually $5.6 \ dm^3$ at $\text{STP},$ calculate the number of molecules in the actual volume of $\text{D}$ at $\text{STP}$.
Answer
| Gases $A$ |
$D$ |
| $1 :4$ |
|
| $5.6\ dm ^3$ |
$4 \times 5.6 \ dm ^3$ at $\text{STP}$
$22.4\ dm ^3 ($molar volume$)$
$6 \times 10^{23}$ molecules. |
View full question & answer→Question 203 Marks
The equations given below relate to the manufacture of sodium carbonate $($Molecular weight of $Na_2CO_3 = 106).$
$NaCl + NH_3 + CO_2 + H_2O \rightarrow NaHCO_3 + NH_4Cl$
$2NaHCO_3\rightarrow Na_2CO_3 + H_2O + CO_2$
What mass of sodium hydrogen carbonate must be heated to give $21.2g$ of sodium carbonate? based on the production of $21.2g$ of sodium carbonate.
Answer$2NaHCO_3\rightarrow Na_2CO_3 + H_2O + CO_2$
Molecular mass of $NaHCO_3 = 84$
Molecular mass of $Na_2CO_3 = 106$
From the above reaction :
$1 Na_2CO_3$ is obtained from $= 2NaHCO_3$
$106g$ of $Na_2CO_3$ is obtained from $= 168g$ of $NaHCO_3$
So$, 21.2 g$ of $Na_2CO_3$ will be obtained from $= 168 \times 21.2/106 = 33.6g$ of $NaHCO_3$
View full question & answer→Question 213 Marks
When heated, potassium permanganate decomposes according to the following equation $:2KMnO_4 \rightarrow K_2MnO_4+ MnO_2 + O_2$
Given that the molecular mass of potassium permanganate is $158g,$ what volume of oxygen $($measured at room temperature$)$ would be obtained by the complete decomposi tion of $15.8g$ of potassium permanganate?
$($Molar volume at room temperature is $24$ litres$). [K = 39, Mn = 55, 0 = 16 ]$
Answer$2KMnO_4 \rightarrow K_2MnO_4+ MnO_2 + O_2$
Molecular mass of $K_2MnO_4 = 158$
Molar volume of $O_2$ at room temperature $= 24L$
$2 \times 158g$ of $K_2MnO_4$ at room temperature yields $= 24L$ of $O_2$
$15.8g$ of $K_2MnO_4$ will yield $= 24 \times 15.8/2 \times 158 = 1.2L$ of $O_{2.}$
View full question & answer→Question 223 Marks
When heated, potassium permanganate decomposes according to the following equation $:2KMnO_4 \rightarrow K_2MnO_4+ MnO_2 + O_2$
Some potassium permanganate was heated in test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of $1.32g$ . If one litre of hydrogen under the same conditions of temperature and pressure has a mass of $0.0825g,$ calculate the relative molecular mass of oxygen.
AnswerMass of one litre of oxygen gas liberated at room temperature $= 1.32g$
Mass of one litre of hydrogen under the same conditions of temperature and pressure $= 0.0825g$
Relative Molecular mass of oxygen $=$ Weight of n molecule of $O_2$ Weight of $1/2$ molecule of hydrogen
$= 1.32 \times 2/ 0.0825 = 32\ a.m.u.$
View full question & answer→Question 233 Marks
What volume of oxygen is required to burn completely a mixture of $22.4 \ dm^3$ of methane and $11 .2 \ dm^3$ of hydrogen in to carbon dioxide and steam? Equations of the reactions are given below:
$CH_4 + 2O_2\rightarrow CO_2 + 2H_2O$
$2H_2 + O_2\rightarrow 2H_2O$
Answer$CH_4 + 2O_2\rightarrow CO_2 + 2H_2O$
$1\ vol. 2 \ vol. 1\ vol. 2\ vol.$
One volume of methane requires oxygen $= 2 \ vol.$
So$, Vol.$ of oxygen used $= 2 \times 22.4 = 44.8 \ dm^3$
$2H_2 + O_2 \rightarrow 2H_2O$
$2 \ vol. 1 \ vol. 2 \ vol.$
$2$ volume of hydrogen needs one volume of oxygen
So, Volume of oxygen used $= 22.4 / 2 = 11.2\ dm^3$
Total volume of oxygen used $= 44.8 + 11.2 = 56.0\ dm^3$
View full question & answer→Question 243 Marks
Calculate the percentage of platinum in ammonium chloroplatinate $(NH_4)_2PtCl_6.$
$[N = 14, H = 1, Pt = 195, Cl =35.5]$
$($Give your answer correct to the nearest whole number$)$
AnswerMolecular formula of ammonium chloroplatinate $(NH_4)_2PtCl_6:$
$2 \times ($atomic mass of $N + 8 \times$ atomic mass of $H) +$ atomic mass of platinum $+ \ 6 \ \times$ atomic mass of chlorine
$2 \times (14 + 8 ) + 195 + 6 \times 35.3= 444$
$444$ parts of ammonium chloroplatinate contains $195$ parts by weight of platinum
So$, 100$ parts will contain $= 195 \times 100/444 = 43.9\% = 44\%$
View full question & answer→Question 253 Marks
The reaction $4N_2O + CH_4 \rightarrow CO_2 + 2H_2O + 4N_2$ takes place in the gaseous state. If the volumes of all the gases are measured at the same temperature, and pressure, calculate the volume of dinitrogen oxide $(N_2O),$ required to give $150\ cm^3$ of steam.
AnswerAccording to Gay$-$Lussac's law : In the equation
$4N_2O + CH_4\rightarrow CO_2 + 2H_2O + 4N_2$
$Vol.$ of $H_2O$ produced is $= 2 \ vol. = 150\ cm^3$
$Vol.$ of $N_2O$ required is $= 4\ vol. = 150 \times 4 / 2 = 300 \ cm^3$
$300\ cm^3$ of dinitrogen oxide $(N_2O)$ is required to give $150 \ cm^3$ of steam.
View full question & answer→Question 263 Marks
What volume of oxygen would be required for complete combustion of $100L$ of ethane according to the following equation?
$2C_2H_6 + 7O_2\rightarrow 4CO_2 + 6H_2O$
AnswerMolecular mass of ethane $= 30$
According to Gay$-$Lussac's law :
$2 \ vol.$ of $C_2H_6$ requires $= 7 \ vol.$ of oxygen
$Vol.$ of $C_2H_6 = 2 \ vol. = 100 L$
$Vol.$ of oxygen required $= 7 \ vol. = 350L$
View full question & answer→Question 273 Marks
When excess lead nitrate solution was added to a solution of sodium sulphate$, 15.1g$ of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
$Na_2SO_4 + Pb(NO_3)_2 \rightarrow PbSO_4 + 2NaNO_3$
$(H = 1, C = 12, O = 16, Na = 23, S = 32, Pb = 207)$
Answer$Na_2SO_4 + Pb(NO_3)_2 \rightarrow PbSO_4 + 2NaNO_3$
molecular weight of $Na_2SO_4$ is $142g$
molecular weight of $PbSO_4$ is $303g$
$303g$ of $PbSO_4$ is formed by $142g$ of $Na_2SO_4$
$15.1g$ of $PbSO_4$ is formed by $= 142 \times 15.1/303 = 7.1g$ of $Na_2SO_{4.}$
View full question & answer→Question 283 Marks
Washing soda has the formula $Na_2CO_3.10H_2O$.What is the mass of anhydrous sodium carbonate left when all the water of crystallization is expelled by heating $57.2 g$ of washing soda?
Answer$Na _2 CO _3 \cdot 10 H _2 O \xrightarrow{\Delta} Na _2 CO _2+10 H _2 O$
molecular weight of washing soda is $286.14g$
molecular weight of sodium carbonate is $106g$
$286.14g$ of $Na_2CO_3. 10H_2O$ forms $106g$ of sodium carbonate on heating $57.2 g$ of $Na_2CO_3. 10H_2O$ forms $= 106 \times 57.2/286.14 = 21.2g$
View full question & answer→Question 293 Marks
If the hydrogen sample contains N molecules, how many molecules are present in oxygen sample?
AnswerAmount of hydrogen and oxygen gases is same $=2 g$
So, for oxygen $32 g$ of gas has $= N$ molecules
Then, $2 g$ of gas has $= N / 32 \times 2=16$
No of molecules of oxygen $= N / 16$.
View full question & answer→Question 303 Marks
Each of the two flasks contains $2.0 g$ of gas at the same temperature and pressure. One flask contains oxygen and the other hydrogen. Which sample contains the greater number of molecules?
AnswerFlask having oxygen gas :
$16g$ of oxygen gas has $= 6.023 \times 10^{23}$ molecules.
$2g$ of oxygen gas will have $= 6.023 \times 10^{23}/16 \times 2 = 0.75 \times 10^{23}$
Flask having hydrogen :
$1g$ of hydrogen has $= 6.023 \times 10^{23}$ molecules.
$2g$ of hydrogen gas will have $= 6.023 \times 10^{23} \times 2/1 = 12.05 \times 10^{23}$
So, hydrogen gas has greater number of molecules.
View full question & answer→Question 313 Marks
A vessel contains N molecules of oxygen at a certain temperature and pressure. How many molecules of sulphur dioxide can the vessel accommodate at the same temperature and pressure?
AnswerAs we know from Avogadro's law that under same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules so if a vessel contains N molecules of oxygen at a certain temperature and pressure and since the vessel remains same so same volume of sulphur dioxide will be present in the vessel and hence N molecules of sulphur dioxide can be accommodated in the vessel at the same temperature and pressure.
View full question & answer→Question 323 Marks
If a crop of wheat removes $20 \ Kg$ of nitrogen per hectare of soil, what mass of the fertilizer calcium nitrate$,Ca(NO_3)_2$ would be required to replace nitrogen in $10$ hectare field? $(N = 14, O = 16, Ca = 40)$
AnswerMolecular weight of $Ca(NO_3)_2 = 164$
$164$ parts by weight of calcium nitrate contains $28$ parts by weight of nitrogen .
$28 \ Kg$ of nitrogen will be replaced by $= 164 \ Kg$ of $Ca(N0_3)_2$
$20 \ Kg$ of nitrogen will be replaced by $= 164 \times 20/28$
$= 117.14 \ kg$
For$, 1$ hectare of field $20\ Kg$ of nitrogen will be replaced by $= 117.14 \ Kg$ of $Ca(NO_3)_2$
For$, 10$ hectare of field $20\ Kg$ of nitrogen will be replaced by $= 117.14 \times 10 = 1171.4\ Kg$
Hence$, 1171.4 \ Kg$ of the fertilizer calcium nitrate$,Ca(NO_3)_2$ would be required to replace nitrogen in $10$ hectare field .
View full question & answer→Question 333 Marks
Ammonia may be oxidized to ni trogen monoxide in the presence of a catalyst according to th e equation as:
$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$
If $27L$ of reactants are consumed, what volume of ni trogen monoxide is produced at the same temperature and pressure?
Answer$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$
$4\ vol.\ 5 \ vol.\ 4\ vol.$
Volume of reactants $= 4\ vol.$ of ammonia $+ \ 5\ vol.$ of oxygen $= 9\ vol. $
$9\ vol.$ of reactants produces $4\ vol.$ of Nitric oxide
Therefore$, 27\ vol.$ of reactants will produce $4 \times 27$ liter $=$
$12$ litres of Nitric oxide.
View full question & answer→Question 343 Marks
Concentrated nitric acid oxidizes phosphorous to phosphoric acid according to the following equation :
$P + 5HNO_3 \rightarrow H_3PO_4 + H_2O + 5NO_2$
What mass of phosphoric acid can be prepared from $6 .2 g$ of phosphorous?
AnswerGiven equation is :
$P + 5HNO_3 \rightarrow H_3PO_4 + H_2O + 5NO_2$
Molecular mass of phosphorous $= 31$
Molecular mass of phosphoric acid $= 98$
$31g$ of phosphorous produces $= 98g$ of phosphoric acid
$6.2g$ of phosphorous will produce $= 98 \times 6.2/31 = 19.6g$
Hence$, 19.6g$ of phosphoric acid can be prepared from $6.2g$ of phosphorous.
View full question & answer→Question 353 Marks
Water can split into hydrogen and oxygen under suitable conditions. The equations representing the change is:
$2H_2O(I) \rightarrow 2H_2 (g) + O_2(g)$
Ammonia burns in oxygen and the combustion in the presence of a catalyst may be represented as:
$2NH_3 (g) +2^{1/2}O_2 (g) \rightarrow 2NO (g) + 3H_2O (I)$
What mass of steam is produced when $1.5 g$ of nitrogen monoxide is formed?
AnswerGiven equation is:
$2NH_3 (g) +2^{1/2}O_2 (g) \rightarrow 2NO (g) + 3H_2O (I)$
Molecular mass of $NO = 30$
Molecular mass of $H_2O = 18$
From the equation :
$2 \ moles$ of $NO = 3 \ moles$ of $H_2O$
$60g$ of $NO = 54g$ of $H_2O.$
$1.5g$ of $NO = 54 \times 1.5 / 60 = 1.35g$ of $H_2O.$
View full question & answer→Question 363 Marks
Water can split into hydrogen and oxygen under suitable conditions. The equations representing the change is:
$2H_2O(I) \rightarrow 2H_2 (g) + O_2(g)$
If a given experiment results in $2500\ cm^3$ of hydrogen being produced, what volume of oxygen is liberated at the same time under the same conditions of temperature and pressure?
AnswerGiven equation is: $2H_2O (I) \rightarrow 2H_2 (g) + O_2 (g)$
According to Gay$-$Lussac's law;
$2$ volume of water produces $2$ volume of hydrogen and $1$ volume of oxygen
i.e $2$ volume of water produces $= 2$ volume of hydrogen
$= 2500\ cm^3$
$2$ volume of water will produce $=1$ volume of Oxygen
$= 2500/ 2 = 1250 \ cm^3$
i.e $0= 1250 \ cm^3$
View full question & answer→Question 373 Marks
If the relative molecular mass of A is 90, what is the molecular formula of A?
AnswerThe relative molecular mass of $A$ is 90
Empirical Formula mass of compound is $=45$
Then, $n =$ Molecular mass/ Empirical Formula Mass
$=90 / 45=2$
Molecular formula of compound $= n x$ Empirical formula
$=2 \times\left( CO _2 H \right)= C _2 O _4 H _2 .$
View full question & answer→Question 383 Marks
Is it possible to change the temperature and pressure of a fixed mass of a gas without changing its volume? Explain your answer.
AnswerNo. it is not possible to change the temperature and pressure of a fixed mass of a gas without changing its volume because all the three variables are interrelated to each other by the gas equation as :
PV/T = K (constant) -----1)
Hence if we change any one or two of the variables in the above equation then automatically third variable also has to change to make equation 1 equal to a constant.
View full question & answer→Question 393 Marks
Potassium nitrate on strong heating decomposes as under :
$2KNO_3\rightarrow 2KNO_2 + O_2$
Calculate : Weight of oxygen formed when $5.05g$ of potassium nitrate decomposes completely.
$(K = 39, 0 = 16, N = 14)$
AnswerGiven equation is $: 2KNO_3 \rightarrow 2KNO_2 + O_2$
Molecular mass of $KNO_3$ is $: ($Atomic mass of $K\ +$ Atomic mass of $N\ +$ Atomic mass of $O) = (39 + 14 + 16 x 3 ) = 101$
Molecular mass of of $KNO_2 = (39 + 14 + 16 \times 2) = 85$
Now, decomposition of $101g$ of $KNO_3$ yield $= 16g$ of $O_2$
So, decomposition of $5.05 g$ of $KNO_3$ will yield $= 16 \times 5 .05 / 101 = 0.8 g$
Hence, when $5.05g$ of potassium nitrate decomposes completely $0.8 g$ of oxygen is formed.
View full question & answer→Question 403 Marks
Empirical formula of a compound is $CH_2O.$ If its empirical formula is equal to its vapour density, calculate the molecular formula of the compound.
AnswerEmpirical formula of the compound is $CH_2O.$
Empirical formula mass $=$ Atomic mass of $C\ +$ Atomic mass of $H\ +$ Atomic mass of $O$
$= 12 + 2 \times 1 + 16 = 30.$
Now as empirical formula is equal to the vapour density then;
Molecular mass $= 2 \ \times$ vapour density
$= 2 \times 30 = 60$
n $=$ Molecular mass / Empirical formula mass
$= 60 / 30 = 2$
Molecular formula $=$ n \ times empirical formula
$= 2 \times (CH_2O) = C_2H_4O_2$
The molecular formula of the compound is $C_2H_4O_2$
View full question & answer→Question 413 Marks
Calculate the percentage composition of oxygen in lead nitrate $[Pb(NO>3)2]. [Pb = 207, N= 14, O = 16]$
Answer$Pb + (N)_2 + (O)_6$
$207 + 2 \times 14 + 6 \times 16 = 331.$
So, the molecular mass of $Pb(NO_3)_2 = 331.$
$331$ by weight of $Pb(NO_3)_2$ contain $96$ parts by weight of oxygen .
$100$ parts will contain $= 96 \times 100 / 331 = 29 \%$
So, the percentage composition of oxygen in lead nitrate is $29\%$
View full question & answer→Question 423 Marks
Calculate the number of molecules in $12.8g$ of sulphur dioxide gas. Take Avogadro's number as $6 \times 10^{23}.$
$[S = 32, 0 = 16 ]$
AnswerNumber of molecules in $12.8g$ of sulphur dioxide gas.
Molecular mass of $SO_2 = 64\ a.m.u .$
So$, 64g = 1 \ mole$
$12.8g = 12.8/ 64 = 0.2 \ mole$
Now $1\ mole$ of $SO_2$ contains $= 6 \times 10^{23}$ molecules
$0 .2 \ mole$ of $SO_2$ contains $= 0 .2 \times 6 \times 10^{23}$
$= 1.2 \times 10^{23}$ molecules.
View full question & answer→Question 433 Marks
Nitrogen and oxygen gas react as illustrated by the equation given below $: N_2 + O_2 \rightarrow 2NO$
Calculate the volume of each reacting gas required to produce $1400\ cm^3$ of nitric oxide.
AnswerGiven react ion is :
$N_2 + O_2 \rightarrow 2NO$
According to Gay $-$ Lussac's law in the above reaction $1$ volume of nitrogen combines with $1$ volume of oxygen to produce $2$ volumes of nitric oxide.
i .e. $N_2 + O_2 \rightarrow 2NO$
$1 \ vol. 1 \ vol. 2 \ vol.$
The volume of nitric oxide produced is $= 1400\ cm^3.$
Let the volumes of nitrogen and oxygen gases be $= x$
Then$, N_2 + O_2\rightarrow 2NO$
$ x \ \times \ 1400\ cm3$
So$, x + x = 1400$
$2x = 1400$
$x= 1400/ 2 = 700\ cm^3$
Hence the volumes of reacting gases i .e. nitrogen and oxygen is $700 \ cm^3$ each.
View full question & answer→Question 443 Marks
A student puts his signature with graphite pencil. If the mass of carbon in the signature is $10^{-12} g$. Calculate the number of carbon atoms in the signature.
Answer$12g$ of carbon contains $= 6.023 \times 10^{23}$ number of carbon atoms.
$10^{-12}g$ of carbon will contain $=\frac{6.023 \times 10^{23} \times 10^{-12}}{12}$
$= 0.5019 \times 10^{11}$
$= 5.019 \times 10^{10}$
So,the number of carbon atoms in the signature is $5.019 \times 10^{10}$
View full question & answer→Question 453 Marks
$10g$ of $NaCl$ solution is mixed with $17g$ of silver nitrate solution. Calculate the weight of silver chloride precipitated.
$AgNO_3 + NaCl \rightarrow AgCI + NaNO_3$
AnswerGiven reaction is :
$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$
Molecular mass of $AgNO_3$ is $170$
Molecular mass of $AgCI$ is $143.5$
$170g$ of $AgNO_3$ produces $= 143.5g$ of $AgCl$
$17g$ of $AgNO_3$ will produce $= 143.5 \times 17/170 = 14.35g$
So, the weight of silver chloride precipitated is $14.35g$
View full question & answer→Question 463 Marks
What weight of sulphuric acid will be required to dissolve $3g$ of magnesium carbonate?
$[Mg = 24, C =12, 0 = 16 ]$
$MgCO_3 + H_2SO_4 \rightarrow MgSO_4 + H_2O+ CO_2$
AnswerMolecular formula of $MgCO_3$ is $= 84$
Molecular formula of $H_2SO_4 = 98$
Now if$, 84g$ of $MgCO_3$ requires $= 98g$ of $H_2SO_4.$
$3g$ of $MgC0_3$ will require $= 98 \times 3/84 = 3.5g$
So$, 3.5g$ of sulphuric acid will be required to dissolve $3g$ of magnesium carbonate.
View full question & answer→Question 473 Marks
Explain the terms empirical formula and molecular formula.
AnswerEmpirical formula:"Empirical formula of a compound is the formula which gives the number of atoms of different elements present in one molecule of the compound, in the simplest numerical ratio".
Molecular formula: "Molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the compound".
View full question & answer→Question 483 Marks
The mass of 11.2 litres of a certain gas at s.t.p is 24 g. Find gram molecular mass of the gas
AnswerGiven :
Volume of gas at STP = 11.2 litres
Mass of gas at STP = 24g
Gram molecular mass = ?
Mass of 22.4L of a gas at STP is equal to its gram molecular mass.
11.2L of the gas at STP weight 24g
So,
22.4L of the gas will weight
$\frac{24}{11.2}$ x 22.4 = 48g
Gram molecular mass = 48g.
View full question & answer→Question 493 Marks
A compound of $X$ and $Y$ has the empirical formula $XY_2.$ Its vapour density is equal to its empirical formula weight. Determine its molecular formula.
AnswerGiven :
Empirical formula $: XY_2$
Vapour density $=$ Empirical formula weight
Molecular formula $= ?$
Molecular weight $= n($ Empirical formula weight$)$
$= 2 \times V.D.$
$n($ Empirical formula weight$) = 2 \times V.D.$
Since, Vapour density $=$ Empirical formula weight
$n = 2$
molecular formula $= 2($Empirical formula$)$
$ = 2(XY_2)$
Molecular formula $= X_4Y_{4.}$
View full question & answer→Question 503 Marks
Calculate the percentage of water of crystallization in $CuSO_{4. }5H_2O$
$(H = 1, O = 16, S = 32, Cu = 64)$
AnswerMolar mass of $CuSO_4.5H_2O = ( 64 ) + ( 32 ) + (9 \times 16) + (10 \times 1) = 250$
Molar mass of $5.H_2O$ molecules $= ( 10 \times 1) + ( 5 \times 16) = 90$
Percentage of water of Crystallisation $= \frac{90 \times 100}{250} = 36\%$
View full question & answer→